1 blue dice and 2 red Dice are thrown....what is the probability that the no. appeared on blue dice is always greater than both the no. appeared on the 2 red dice..

• if the no. on blue dice is 2 then 2 red dice can take (1,1) value = 1^2
  if the no. on blue dice is 3 then 2 red dice can take (1,1)(1,2)(2,1)(2,2) = 2^2
  if the no. on blue dice is 4 then 2 red dice can take (1,1)(1,2)(2,1)(2,2)(1,3)(2,3)(3,2)(3,1)(3,3) = 3^2
  Similarly, if the no. on blue dice is 5 then 2 red dice can take = 4^2
  and if the no. on blue dice is 6 then 2 red dice can take = 5^2
  
  therefor, probability = total no. of favourable outcome/total no. of possible outcome
  = 1^2+2^2+3^2+4^2+5^2 / 216
  = 55/216
  
• 9 years agoHelpfull: Yes(50) No(6)
• what are the most imp topics to cover for clearing tcs wrtitten test ?
• 9 years agoHelpfull: Yes(15) No(3)
• for the number appeared on the blue to be always greater than the both the numbers appeared on the 2 red dice the numbers that can be on blue dice are 3 or 4 or 5 or 6
  1 and 2 cannot be taken because if blue dice is 1 it cant be greater than the two red dice in any case and if blue dice is 2 it cant be greater than the two red dice but can be equal
  if B=2 and R1=1 & R2=1 in this case blue dice is equal to the sum of both red dice but not greater
  
  so clearly the possible cases for blue are 3 4 5 6
  
  if the no. on blue dice is 3 then 2 red dice can take (1,1)(1,2)(2,1)(2,2) = 2^2
  if the no. on blue dice is 4 then 2 red dice can take (1,1)(1,2)(2,1)(2,2)(1,3)(2,3)(3,2)(3,1)(3,3) = 3^2
  Similarly, if the no. on blue dice is 5 then 2 red dice can take = 4^2
  and if the no. on blue dice is 6 then 2 red dice can take = 5^2
  
  therefor, probability = total no. of favourable outcome/total no. of possible outcome
  = 2^2+3^2+4^2+5^2 / 216
  = 54/216
  =1/4
• 9 years agoHelpfull: Yes(10) No(5)
• if one red(1) second red(1) then blue can(2,3,4,5,6)=(1/6)(1/6)(5/6) +
  ,, red(1,2) ,, red(1,2) then blue can(3,4,5,6)=2!(1/6)(1/6)(4/6) +
  ,, red(1,2,3) ,, red(1,2,3) then blue can(4,5,6)=3!(1/6)(1/6)(3/6) +
  ,, =4!(1/6)(1/6)(2/6) +
  ,, =5!(1/6)(1/6)(1/6)
  
  
  =26/27
• 9 years agoHelpfull: Yes(3) No(7)
• For 1 and 2 no on blue dice the no on the two simultaneously thrown red dices will be greater or equal to the 1,2 no on the blue dice.
  for 3 no on blue combinations in red are {1,1}
  for 4 no on blue combinations in red are {1,1},{1,2},{2,1}
  for 5 no on blue dice combinations in red are {1,1},{1,2},{1,3},{2,1},{2,2},{3,1}
  for 6 no on blue dice com in red are {1,1},{1,2},{1,3},{1,4},{1,5},{2,1},2,2},{2,3},{3,1},{3,2},{4,1}
  so probability is 21/216
• 9 years agoHelpfull: Yes(2) No(3)
• 1)when blue dice has number 2,,then(1,1).....1 way
  2)when blue dice has number 3,,then(1,1),(21),(12) ...3 ways
  3)when blue dice has number 4,,then(1,1),(21),(12),(22),(31),(13)...6 ways
  4)when blue dice has number 5,,then(1,1),(21),(12),(22),(31),(13),(32),(23),(41),(14),...10 ways
  5)when blue dice has number6,then(1,1),(21),(12),(22),(31),(13),(32),(23),(41),(14),(24),(42),(51),(15),(33)....15ways.....
  thos prob=35/216
• 9 years agoHelpfull: Yes(1) No(0)
• 
  when on blue 1then possible outcome is 25
  when blue 2 then the possible outcome is 16
  when blue is 3 then the possible outcome is 9
  when blue is 4 then the possible outcome is 4
  when blue is 5 then the possible outcome is 1
  
  p=55/216
• 9 years agoHelpfull: Yes(1) No(0)